How is this concise partial derivative statement true?

Question

If $xe^z=y^2z $ then $$\frac{\partial z}{\partial x}=\frac{e^z}{y^2-xe^z}$$

This question has, honestly, stumped me.

Answer 1

Differentiate both sides implicitly to get $$ \frac{\partial \left[xe^z\right]}{\partial x} = e^z + xe^z z' $$ and $$ \frac{\partial \left[y^2 z\right]}{\partial x} = y^2z' $$ where $z' = \frac{\partial z}{\partial x}$, assuming $y$ is constant wrt $x$.

Can you finish?

Answer 2

The equation can be rewritten as,

$$z = \frac{xe^z}{y^2}$$

Differentiating partially w.r.t $x$ and applying chain rule,

$$\frac{\partial z}{\partial x}= \frac{1}{y^2}\left(e^z+xe^z\frac{\partial z}{\partial x}\right) \implies \frac{\partial z }{\partial x}(y^2-xe^z) = e^z \implies\color{blue}{\frac{\partial z }{ \partial x} = \frac{e^z}{y^2 - xe^z}}$$